Let the equation x(x + 2)(12 - k) = 2$x(x + 2)(12 - k) = 2$ have equal roots. Then the distance of the point (k, frack2)$(k, \frac{k}{2})$ from the line 3x + 4y + 5 = 0$3x + 4y + 5 = 0$ is

### Related Formula textSubstitute variable to convert non-linear form: alpha = frac1sqrtx quad (x > 0)$$\text{Substitute variable to convert non-linear form: } \alpha = \frac{1}{\sqrt{x}} \quad (x > 0)$$ ### Core Logic Let frac1sqrtx = alpha$\frac{1}{\sqrt{x}} = \alpha$. The equation reduces to a product of two quadratics: (9alpha^2 - 9alpha + 2)(2alpha^2 - 7alpha + 3) = 0$$(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) = 0$$ ### Step 1: Factorize the components First quadratic: 9alpha^2 - 9alpha + 2 = 0 implies (3alpha - 2)(3alpha - 1) = 0 implies alpha = frac23, frac13$9\alpha^2 - 9\alpha + 2 = 0 \implies (3\alpha - 2)(3\alpha - 1) = 0 \implies \alpha = \frac{2}{3}, \frac{1}{3}$ Second quadratic: 2alpha^2 - 7alpha + 3 = 0 implies (2alpha - 1)(alpha - 3) = 0 implies alpha = frac12, 3$2\alpha^2 - 7\alpha + 3 = 0 \implies (2\alpha - 1)(\alpha - 3) = 0 \implies \alpha = \frac{1}{2}, 3$ ### Step 2: Solve for x Since alpha = frac1sqrtx implies x = frac1alpha^2$\alpha = \frac{1}{\sqrt{x}} \implies x = \frac{1}{\alpha^2}$. For alpha = frac13 implies x = 9$\alpha = \frac{1}{3} \implies x = 9$ For alpha = frac12 implies x = 4$\alpha = \frac{1}{2} \implies x = 4$ For alpha = frac23 implies x = frac94$\alpha = \frac{2}{3} \implies x = \frac{9}{4}$ For alpha = 3 implies x = frac19$\alpha = 3 \implies x = \frac{1}{9}$ All 4 values are positive and valid. ### Pattern Recognition Always check constraints first (x > 0$x > 0$ due to sqrtx$\sqrt{x}$ in denominator). Since all roots alpha > 0$\alpha > 0$, every single algebraic root maps to a real distinct solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations

### Related Formula Conjugate Surd Identity: (sqrta + sqrtb)(sqrta - sqrtb) = a - b$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$ ### Core Logic Observe the base components of the exponents: (sqrt3 + sqrt2)(sqrt3 - sqrt2) = 3 - 2 = 1$$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$ Therefore, we can express one base as the reciprocal of the other: sqrt3 - sqrt2 = frac1sqrt3 + sqrt2$$\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}$$ The equation becomes: (sqrt3 + sqrt2)^x + frac1(sqrt3 + sqrt2)^x = 10 $$(\sqrt{3} + \sqrt{2})^{x} + \frac{1}{(\sqrt{3} + \sqrt{2})^{x}} = 10 $$ ### Step 1: Formulate the Quadratic Equation Let (sqrt3 + sqrt2)^x = t$(\sqrt{3} + \sqrt{2})^{x} = t$. Then: t + frac1t = 10 implies t^2 - 10t + 1 = 0 $$t + \frac{1}{t} = 10 \implies t^{2} - 10t + 1 = 0 $$ Solving for t$t$ using the quadratic formula: t = frac10 pm sqrt(-10)^2 - 4(1)(1)2 = frac10 pm sqrt962 = 5 pm 2sqrt6 $$t = \frac{10 \pm \sqrt{(-10)^{2} - 4(1)(1)}}{2} = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6} $$ ### Step 2: Solve for x Notice that (5 pm 2sqrt6)$(5 \pm 2\sqrt{6})$ can be written as square powers of the original base: (sqrt3 pm sqrt2)^2 = 3 + 2 pm 2sqrt6 = 5 pm 2sqrt6$$(\sqrt{3} \pm \sqrt{2})^{2} = 3 + 2 \pm 2\sqrt{6} = 5 \pm 2\sqrt{6}$$ Thus, we have: - For t = 5 + 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 + sqrt2)^2 implies x = 2$t = 5 + 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} + \sqrt{2})^{2} \implies x = 2$ - For t = 5 - 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 - sqrt2)^2 = (sqrt3 + sqrt2)^-2 implies x = -2$t = 5 - 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} - \sqrt{2})^{2} = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2$ Therefore, the distinct real solutions are x = 2$x = 2$ and x = -2$x = -2$. The number of elements in set S$S$ is 2$2$. ### Pattern Recognition Sees: Rational conjugate bases added with inverse matching variables. Shortcut: Whenever you see an equation of the form A^x + B^x = C$A^x + B^x = C$ where AB = 1$AB = 1$, the solution will always be symmetric (pm x_0$\pm x_0$). Checking x=2$x=2$ explicitly gives (sqrt3+sqrt2)^2 + (sqrt3-sqrt2)^2 = (5+2sqrt6) + (5-2sqrt6) = 10$(\sqrt{3}+\sqrt{2})^2 + (\sqrt{3}-\sqrt{2})^2 = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$, confirming pm 2$\pm 2$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 9 Mathematics: Number Systems (Rationalization)

### Related Formula textZero Product Property: A cdot B = 0 implies A = 0 text or B = 0$$\text{Zero Product Property: } A \cdot B = 0 \implies A = 0 \text{ or } B = 0$$ ### Core Logic Given equation: xleft(x^2 + 3|x| + 5|x - 1| + 6|x - 2|right) = 0$$x\left(x^2 + 3|x| + 5|x - 1| + 6|x - 2|\right) = 0$$ This factors into two possibilities: 1) x = 0$x = 0$ 2) x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0$x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0$ ### Step 1: Evaluating the Modulus Term Look at the second factor: f(x) = x^2 + 3|x| + 5|x - 1| + 6|x - 2|$f(x) = x^2 + 3|x| + 5|x - 1| + 6|x - 2|$. Notice that all terms inside are strictly non-negative: - x^2 ge 0$x^2 \ge 0$ - 3|x| ge 0$3|x| \ge 0$ - 5|x - 1| ge 0$5|x - 1| \ge 0$ - 6|x - 2| ge 0$6|x - 2| \ge 0$ For the sum to be 0$0$, every single term must be simultaneously zero. x^2 = 0 implies x = 0$x^2 = 0 \implies x = 0$ However, if x = 0$x = 0$, then 5|x-1| = 5(1) = 5 neq 0$5|x-1| = 5(1) = 5 \neq 0$. Therefore, there is no real value of x$x$ that makes this entire second factor equal to zero. ### Step 2: Conclusion The only valid solution to the equation is x = 0$x = 0$ from the first factor. Thus, there is exactly 1$1$ real solution. ### Pattern Recognition A sum of absolute values and squares set to 0 requires all individual components to hit 0 concurrently. If they have different zero-nodes (0, 1, 2$0, 1, 2$), the sum can never be 0$0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations

### Core Logic Given equation: (a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0$(a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0$. Expand and rearrange into perfect squares: (a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$$(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$$ (ax - b)^2 + (bx - c)^2 = 0$$(ax - b)^2 + (bx - c)^2 = 0$$ Since squares must be non-negative, each term is zero: ax - b = 0 implies x = fracba$$ax - b = 0 \implies x = \frac{b}{a}$$ bx - c = 0 implies x = fraccb$$bx - c = 0 \implies x = \frac{c}{b}$$ Thus, b = ax$b = ax$ and c = bx = ax^2$c = bx = ax^2$. Since a,b,c$a,b,c$ form a triangle, the triangle inequality holds: 1) a + b > c implies a + ax > ax^2 implies x^2 - x - 1 < 0 implies frac1-sqrt52 < x < frac1+sqrt52$a + b > c \implies a + ax > ax^2 \implies x^2 - x - 1 < 0 \implies \frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$ 2) a + c > b implies a + ax^2 > ax implies x^2 - x + 1 > 0$a + c > b \implies a + ax^2 > ax \implies x^2 - x + 1 > 0$ (Always true for real x$x$) 3) b + c > a implies ax + ax^2 > a implies x^2 + x - 1 > 0 implies x > frac-1+sqrt52$b + c > a \implies ax + ax^2 > a \implies x^2 + x - 1 > 0 \implies x > \frac{-1+\sqrt{5}}{2}$ or x < frac-1-sqrt52$x < \frac{-1-\sqrt{5}}{2}$. Taking the intersection (and noting x > 0$x > 0$ since sides are positive): fracsqrt5-12 < x < fracsqrt5+12$$\frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$$ So, alpha = fracsqrt5-12$\alpha = \frac{\sqrt{5}-1}{2}$ and beta = fracsqrt5+12$\beta = \frac{\sqrt{5}+1}{2}$. Calculate 12(alpha^2 + beta^2)$12(\alpha^2 + \beta^2)$: 12 left( frac6-2sqrt54 + frac6+2sqrt54 right) = 12 left( frac124 right) = 36$$12 \left( \frac{6-2\sqrt{5}}{4} + \frac{6+2\sqrt{5}}{4} \right) = 12 \left( \frac{12}{4} \right) = 36$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

### Related Formula textSum of coefficients = 0 implies x = 1 text is a root.$$\text{Sum of coefficients} = 0 \implies x = 1 \text{ is a root.}$$ ### Core Logic Given f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$. Substituting x = 1$x = 1$: f(1) = a + b - 2c + b + c - 2a + c + a - 2b = 0$$f(1) = a + b - 2c + b + c - 2a + c + a - 2b = 0$$ Thus, one root is 1$1$. Let the other root be alpha$\alpha$. ### Step 1: Find the other root Product of roots = fracc + a - 2ba + b - 2c$\frac{c + a - 2b}{a + b - 2c}$. Since one root is 1$1$, we have: alpha cdot 1 = fracc + a - 2ba + b - 2c$$\alpha \cdot 1 = \frac{c + a - 2b}{a + b - 2c}$$ alpha = fracc + a - 2ba + b - 2c$$\alpha = \frac{c + a - 2b}{a + b - 2c}$$ ### Step 2: Analyze Statement (I) If -1 < alpha < 0$-1 < \alpha < 0$: -1 < fracc + a - 2ba + b - 2c < 0$$-1 < \frac{c + a - 2b}{a + b - 2c} < 0$$ This implies b > fraca + c2$b > \frac{a + c}{2}$ and b + c < 2a$b + c < 2a$. Therefore, b$b$ cannot be the Geometric Mean of a$a$ and c$c$. Statement (I) is true. ### Step 3: Analyze Statement (II) If 0 < alpha < 1$0 < \alpha < 1$: 0 < fracc + a - 2ba + b - 2c < 1$$0 < \frac{c + a - 2b}{a + b - 2c} < 1$$ This gives b > c$b > c$ and b < fraca + c2$b < \frac{a + c}{2}$. Therefore, b$b$ may be the Geometric Mean between a$a$ and c$c$. Statement (II) is true. ### Pattern Recognition When coefficients in a quadratic equation are cyclic and sum to 0, one root is always 1$1$. The other root is directly c/a$c/a$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations Class 11 Maths: Sequences and Series